If |G| is a power of p then every subgroup, including the powers of x, has order dividing |G|, hence G is a p group. Conversely, if some other prime q divides |G|, then G contains a q cycle, and is not a p group. For finite groups, p group means |G| is a power of p.
A sylow subgroup is a maximal subgroup that is also a p group. Use zorn's lemma to show such a group exists when the containing group is infinite.
Let the p group G have a normal subgroup H. Since x*H/x is H, G acts on the set H by conjugation. This action leaves the center of G fixed. When restricted to H, it fixes H intersect the center of G. By the fixed point principle, this fixed set has size n, where n = |H| mod p. Of course H is a p group, so n = 0. The center of G, intersected with H, is always divisible by p.
As a special case, set H = G, and p divides the number of elements in the center of G. Therefore the center of every p group is nontrivial.